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3 years ago

If the current of a circuit is 1.5 A, and the power is 24 W. what is the resistor?

Physics

1 answer:

Ymorist [56]3 years ago

3 0

Answer:

The resistor has a resistance of 10.667 ohms.

Explanation:

By Ohm's Law, voltage ( $V$ ), in volts, is directly proportional to the current ( $i$ ), in amperes, and by definition of power ( $\dot W$ ), in watts, we have the following formula:

$\dot W = i^{2}\cdot R$ (1)

Where $R$ is the resistance, in ohms.

If we know that $\dot W = 24\,W$ and $i = 1.5\,A$ , then the resistance of the resistor is:

$R = \frac{\dot W}{i^{2}}$

$R = 10.667\,\Omega$

The resistor has a resistance of 10.667 ohms.

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3 years ago

Read 2 more answers

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

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3 years ago

Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?

eduard

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = $\frac{1}{2}$ m $v^{2}$

Plug in the numbers = $\frac{1}{2}$ (4.0)( $8^{2}$ )

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

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3 years ago